ELEC Digital Logic Circuits Fall 2015 Boolean Algebra (Chapter 2) Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL Fall 2015, Sep ELEC Lecture 3 1

Binary Arithmetic Switching Theory Semiconductor Technology Digital Systems Fall 2015, Sep 14...ELEC Lecture 32 Boolean Algebra DIGITAL CIRCUITS

George Boole, Born, Lincoln, England Professor of Math., Queen’s College, Cork, Ireland Book, The Laws of Thought, 1853 Wife: Mary Everest Boole Fall 2015, Sep ELEC Lecture 3 3

An Axiom or Postulate A self-evident or universally recognized truth. An established rule, principle, or law. A self-evident principle or one that is accepted as true without proof as the basis for argument. A postulate – Understood as the truth. Fall 2015, Sep ELEC Lecture 3 4

Boolean Algebra Postulate 1: Set and Operators Define a set K containing two or more elements. Define two binary operators: +, also called “OR” ·, also called “AND” Such that for any pair of elements, a and b in K, a + b and a·b also belong to K Fall 2015, Sep ELEC Lecture 3 5

Example a: {students in “digital circuits” course} b: {students in “computer systems” course} 1: {all EE juniors} 0: {null set} K: {a, b, 1, 0, a+b, a·b} Postulate 1: a + b = 1 a · b = {full-time EE students} Fall 2015, Sep ELEC Lecture 3 6

Postulate 2: Identity Elements There exist 0 and 1 elements in K, such that for every element a in K a + 0 = a a · 1 = a Definitions: 0 is the identity element for + operation 1 is the identity element for · operation Remember, 0 and 1 here should not be misinterpreted as 0 and 1 of ordinary algebra. Fall 2015, Sep ELEC Lecture 3 7

Postulate 3: Commutativity Binary operators + and · are commutative. That is, for any elements a and b in K: a + b = b + a a · b = b · a Example: a + b = b + a = {all EE students} a · b = b · a = {all full-time EE students} Fall 2015, Sep ELEC Lecture 3 8

Postulate 4: Associativity Binary operators + and · are associative. That is, for any elements a, b and c in K: a + (b + c) = (a + b) + c a · (b · c) = (a · b) · c Example: EE department has three courses with student groups a, b and c Fall 2015, Sep ELEC Lecture 3 9 All EE students: a + (b + c) a b c ac b EE students in all EE courses: a · (b · c)

Postulate 5: Distributivity Binary operator + is distributive over · and · is distributive over +. That is, for any elements a, b and c in K: a + (b · c) = (a + b) · (a + c) a · (b + c) = (a · b) + (a · c) Remember dot (·) operation is performed before + operation: a + b · c = a + ( b · c) ≠ (a + b) · c Fall 2015, Sep ELEC Lecture 3 10

Postulate 6: Complement A unary operation, complementation, exists for every element of K. That is, for any element a in K: Where,1 is identity element for · 0 is identity element for + Fall 2015, Sep ELEC Lecture 3 11

Example A set contains four elements: x = {φ}, null sety = {1, 2} z = {3, 4, 5}w = {1, 2, 3, 4, 5} Define two operations: union (+) and intersection (·): +xyzw xxyzw yyyww zzwzw wwwww ·xyzw xxxxx yxyxy zxxzz wxyzw Fall 2015, Sep ELEC Lecture 3 12

Verify Postulates 1, 2 and Union and intersection, used as binary operators on a pair of elements, produce a result within the same set Identity elements are x for union (+) and w for intersection (·). x ≡ 0; w ≡ Commutativity is verified from the symmetry in the function tables for the two operators Fall 2015, Sep ELEC Lecture 3 13

Postulate 4: Associativity Examine the Venn diagram. For any group of elements, intersections or unions in any order lead to the same result. Fall 2015, Sep ELEC Lecture 3 14 φ 1, 2 3, 4, 5 x y z w

Postulate 5: Distributivity To verify distributivity, examine the Venn diagram for distributivity over union and intersection. Fall 2015, Sep ELEC Lecture 3 15 φ 1, 2 3, 4, 5 x y z w

Postulate 6: Complements Any element + its complement = Identity for · Any element · Its complement = Identity for + Verifiable from Venn diagram. Fall 2015, Sep ELEC Lecture 3 16 φ 1, 2 3, 4, 5 x y z w Identity Element For + Identity Element For ·

Conclusion Because all six postulates are true for our example, it is a Boolean algebra. Fall 2015, Sep ELEC Lecture 3 17

The Duality Principle Each postulate of Boolean algebra contains a pair of expressions or equations such that one is transformed into the other and vice-versa by interchanging the operators, + ↔ ·, and identity elements, 0 ↔ 1. The two expressions are called the duals of each other. Fall 2015, Sep ELEC Lecture 3 18

Examples of Duals Postulate Duals Expression 1Expression 2 1a, b, a + b ε Ka, b, a · b ε K 2a + 0 = aa · 1 = a 3a + b = b + a a · b = b · a 4a + (b + c) = (a + b) + ca · (b · c) = (a · b) · c 5 a + (b · c)=(a + b) · (a + c)a · (b + c)=(a · b)+(a · c) 6 Fall 2015, Sep ELEC Lecture 3 19

Examples of Duals Fall 2015, Sep ELEC Lecture 3 20 Expressions:A · BA + B Equations: duals A + (BC) = (A+B)(A+C) ↔ A (B+C) = AB + AC Note: A · B is also written as AB. ABA B

Properties of Boolean Algebra Properties stated as theorems. Provable from the postulates (axioms) of Boolean algebra. Fall 2015, Sep ELEC Lecture 3 21

Theorem 1: Idempotency (Invariance) For all elements a in K: a + a = a; a a = a. Proof: a + a=(a + a)1, (identity element) =(a + a)(a + ā),(complement) =a + a ā,(distributivity) =a + 0,(complement) =a, (identity element) Fall 2015, Sep ELEC Lecture 3 22

Theorem 1: Idempotency For all elements a in K: a + a = a; a a = a. Proof: a a=(a a) + 0, (identity element) =(a a) + (a ā), (complement) =a (a + ā),(distributivity) =a 1,(complement) =a, (identity element) Fall 2015, Sep ELEC Lecture 3 23

Theorem 2: Null Elements Exist a + 1 = 1, for + operator. a · 0 = 0, for · operator. Proof: a + 1 =(a + 1)1, (identity element) =1(a + 1),(commutativity) =(a + ā)(a + 1), (complement) =a + ā 1, (distributivity) =a + ā, (identity element) =1, (complement) Similar proof for a 0 = 0. Fall 2015, Sep ELEC Lecture 3 24

Theorem 2: Null Elements Exist a + 1 = 1, for + operator. a · 0 = 0, for · operator. Proof: a + 1 =(a + 1)1, (identity element) =1(a + 1),(commutativity) =(a + ā)(a + 1), (complement) =a + ā 1, (distributivity) =a + ā, (identity element) =1, (complement) Similar proof for a 0 = 0. Fall 2015, Sep ELEC Lecture 3 25

Theorem 2: Null Elements Exist a + 1 = 1, for + operator. a · 0 = 0, for · operator. Proof: a 0=(a 0) + 0, (identity element) =0 + (a 0), (commutativity) =(a ā) + (a 0), (complement) =a(ā + 0), (distributivity) =a ā, (identity element) =0, (complement) Fall 2015, Sep ELEC Lecture 3 26

Theorem 3: Involution Holds a = a Proof: a + ā = 1 and a ā = 0, (complements) or ā + a = 1 and ā a = 0, (commutativity) i.e., a is complement of ā Therefore, a = a Fall 2015, Sep ELEC Lecture 3 27 = =

Theorem 4: Absorption a + a b = a a (a + b) = a Proof: a + a b =a 1 + a b, (identity element) =a(1 + b), (distributivity) =a 1,(Theorem 2) =a,(identity element) Similar proof for a (a + b) = a. Fall 2015, Sep ELEC Lecture 3 28

Theorems 5, 6 and 7 (p ) Theorem 5: Theorem 6: Theorem 7: Fall 2015, Sep ELEC Lecture 3 29

Proving Theorem 5 Fall 2015, Sep ELEC Lecture 3 30 ab Using Venn diagram

Theorem 8: DeMorgan’s Theorem Fall 2015, Sep ELEC Lecture 3 31 Generalization of DeMorgan’s Theorem:

Martians and Venusians Suppose Martians are blue and Venusians are pink. An Earthling identifying itself: “I am not blue or pink.” blue + pink = blue · pink Meaning: “I am not blue and I am not pink.” Or: “I am not a Martian and I am not a Venusian.” Fall 2015, Sep ELEC Lecture 3 32

Tall, Dark and Handsome “He did not appear to be tall, dark and handsome.” tall · dark · handsome = tall + dark + handsome Meaning: “He was not tall or he was not dark or he was not handsome.” Equivalently: “He was short or he was pale or he was ugly.” Perhaps, not the fellow we were looking for. Fall 2015, Sep ELEC Lecture 3 33

Theorem 9: Consensus Fall 2015, Sep ELEC Lecture 3 34 a b c ab āc See page 90. First case for union and intersection: bc

Next, Switching Algebra Set K contains two elements, {0, 1}, also called {false, true}, or {off, on}, etc. Two operations are defined as, + ≡ OR, · ≡ AND. Fall 2015, Sep ELEC Lecture ·